Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #4 : How To Find Increasing Intervals By Graphing Functions

Function A

Graph3

 

 

Function B

Graph4

 

 

 

Function C

Graph5

 

 

Function D

Graph2

 

 

Function E

Graph1

5 graphs of different functions are shown above. Which graph shows an increasing/non-decreasing function?

Possible Answers:

Function E

Function A

Function C

Function B

Function D

Correct answer:

Function E

Explanation:

A function \(\displaystyle f(x)\) is increasing if, for any \(\displaystyle y>x\)\(\displaystyle f(y)\geq f(x)\) (i.e the slope is always greater than or equal to zero)

 

Function E is the only function that has this property. Note that function E is increasing, but not strictly increasing

Example Question #2 : How To Find Increasing Intervals By Graphing Functions

Find the increasing intervals of the following function on the interval \(\displaystyle (0, 5)\):

\(\displaystyle f(x)=x^3-9x\)

Possible Answers:

\(\displaystyle (0, 5)\)

\(\displaystyle (\sqrt{3}, 5)\)

\(\displaystyle (0, \sqrt{3})\)

\(\displaystyle (-\sqrt{3, 0})\cup(\sqrt{3}, 5)\)

Correct answer:

\(\displaystyle (\sqrt{3}, 5)\)

Explanation:

To find the increasing intervals of a given function, one must determine the intervals where the function has a positive first derivative. To find these intervals, first find the critical values, or the points at which the first derivative of the function is equal to zero.

For the given function, \(\displaystyle f'(x)=3x^2-9\).

This derivative was found by using the power rule 

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\)

When set equal to zero, \(\displaystyle x=c=\pm\sqrt{3}\). Because we are only considering the open interval (0,5) for this function, we can ignore \(\displaystyle c=-\sqrt{3}\). Next, we look the intervals around the critical value \(\displaystyle c=\sqrt{3}\), which are \(\displaystyle (0, \sqrt{3})\) and \(\displaystyle (\sqrt{3}, 5)\). On the first interval, the first derivative of the function is negative (plugging in values gives us a negative number), which means that the function is decreasing on this interval. However for the second interval, the first derivative is positive, which indicates that the function is increasing on this interval \(\displaystyle (\sqrt{3}, 5)\)

Example Question #11 : Increasing Intervals

Is \(\displaystyle f(x)\) increasing, decreasing, or flat at \(\displaystyle x=-6\)?

\(\displaystyle f(x)=-3x^5+6x^3\)

Possible Answers:

f(x) is decreasing at the point, because f'(x) is negative.

 f(x) is decreasing at the point, because f'(x) is positive.

f(x) is flat at the point, because f'(x) is zero.

f(x) is increasing at the point, because f'(x) is positive.

Correct answer:

f(x) is decreasing at the point, because f'(x) is negative.

Explanation:

Is f(x) increasing, decreasing, or flat at \(\displaystyle x=-6\)?

\(\displaystyle f(x)=-3x^5+6x^3\)

Recall that to find if a function is increasing or decreasing, we can use its first derivative. If f'(x) is positive, f(x) is increasing. If f'(x) is negative, f(x) is decreasing.

So, given:

\(\displaystyle f(x)=-3x^5+6x^3\)

 

We get

\(\displaystyle f'(x)=-15x^4+18x^2\)

Then:

\(\displaystyle f'(x)=-15(-6)^4+18(-6)^2=-18792\)

Therefore, f(x) is decreasing at the point, because f'(x) is negative.

Example Question #31 : Intervals

Tell whether \(\displaystyle g(t)\) is increasing or decreasing on the interval \(\displaystyle [4,7]\).

\(\displaystyle g(t)=4t^3-3t^2+6t\)

Possible Answers:

Increasing, because g'(t) is negative.

Decreasing, because g'(t) is negative.

Increasing, because g'(t) is positive.

Decreasing, because g'(t) is positive.

Correct answer:

Increasing, because g'(t) is positive.

Explanation:

Tell whether g(t) is increasing or decreasing on the interval [4,7]

\(\displaystyle g(t)=4t^3-3t^2+6t\)

To find increasing and decreasing, find where the first derivative is positive and negative. If g'(t) is positive, then g(t) is increasing and vice-versa.

\(\displaystyle g(t)=4t^3-3t^2+6t\)

\(\displaystyle g'(t)=12t^2-6t+6\)

Then,plug in the endpoints of [4,7] and see what you get for a sign.

\(\displaystyle g'(4)=12(4)^2-6(4)+6=174\)\(\displaystyle g'(7)=12(7)^2-6(7)+6=552\)

So, since g'(t) is positive on the interval, g(t) is increasing.

Example Question #12 : Increasing Intervals

Find the interval on which the function is increasing:

\(\displaystyle f(x)=x^2+x+1\)

Possible Answers:

\(\displaystyle \left(-\frac{1}{2}, \infty \right)\)

The function is never increasing.

\(\displaystyle (-\infty , \infty )\)

\(\displaystyle \left(-\infty , -\frac{1}{2}\right)\)

Correct answer:

\(\displaystyle \left(-\frac{1}{2}, \infty \right)\)

Explanation:

To find the interval(s) on which the function is increasing, we must find the intervals on which the first derivative of the function is positive. 

The first derivative of the function is:

\(\displaystyle f'(x)=2x+1\)

and was found using the rule

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Now we must find the critical value, at which the first derivative is equal to zero:

\(\displaystyle c=-\frac{1}{2}\)

Now, we make the intervals on which we look at the sign of the first derivative:

\(\displaystyle \left(-\infty , -\frac{1}{2}\right), \left(-\frac{1}{2}, \infty \right)\)

On the first interval the first derivative is positive, while on the second it is negative. Thus, the first interval is our answer, because over this range of x values, the first derivative is positive and the function is increasing. 

Example Question #31 : Intervals

Suppose \(\displaystyle f(x)\) is continuous for all \(\displaystyle x\) and known to have at least one root, and \(\displaystyle f'(x)>0\) for all \(\displaystyle x\). Which of the following must be true?

Possible Answers:

\(\displaystyle f\)  has at least one inflection point

\(\displaystyle f\)  has at least one more root

\(\displaystyle f\) has only one root

Correct answer:

\(\displaystyle f\) has only one root

Explanation:

If \(\displaystyle f(x)\) is continuous everywhere and always increasing (i.e. \(\displaystyle f'(x)>0\) for all \(\displaystyle x\)), then it must be true that after \(\displaystyle f\) has attained its root, it can never do so again because it can't "return" to the \(\displaystyle x\)-axis. NOTE: this is not automatically true of functions that aren't continuous. As for the other choices, the possibility of at least having one more root is automatically false and a simple counterexample to the notion thay \(\displaystyle f\) has to have an inflection point is a simple increasing lines. It has a constant positive derivative, but possesses no upward or downward concavity and has no inflection points.

Example Question #15 : Increasing Intervals

Deletable Note to the admin: I am virtually 100% sure the derivative has been correct. Derivative of the top is 6x. Derivative of the bottom is 1/x. So numerator of derivative by quotient rule is \(\displaystyle 6xlnx - 3x^{2}(\frac{1}{x})\). You will note the second term in this is 3x. Denominator is self explanatory. I do not see where it is wrong.

Let   \(\displaystyle y=f(x)=\frac{3x^{2}}{\ln x}\).  On what subintervals of the interval \(\displaystyle (e,\infty )\) is \(\displaystyle f(x)\) increasing?

Possible Answers:

Only intervals where \(\displaystyle lnx\)>\(\displaystyle 3x^{2}\)

\(\displaystyle (e,e^{2})\)

No subinterval of \(\displaystyle (e,\infty )\).

Every nontrivial subinterval of \(\displaystyle (e,\infty )\)

 

 

Correct answer:

Every nontrivial subinterval of \(\displaystyle (e,\infty )\)

 

 

Explanation:

Take the first derivative of \(\displaystyle f(x)\):

 

\(\displaystyle y'= \frac{6xlnx-3x}{ln^{2}x}\)   by quotient rule

 

\(\displaystyle f(x)\) is increasing whenever \(\displaystyle y'\) is positive, that is, whenever both the numerator and denominator are of the same sign. The function \(\displaystyle f(x)=lnx\) is certainly positive for all values of \(\displaystyle x\) greater than \(\displaystyle e\) because  and since 

 

\(\displaystyle \frac{d}{dx}(lnx) = \frac{1}{x}\)

 

is positive for all positive \(\displaystyle x\), it is increasing on the interval, too. It will never be negative. For the same reason, the numerator is always positive. With the numerator and denominator always positive everywhere on the given interval, the derivative is always positive and the function is always increasing. So for any interval of nonzero length within \(\displaystyle (e,\infty )\), \(\displaystyle f(x)\) is increasing.

 

NOTE: Interestingly the opposite of the choice \(\displaystyle f(x)\) > \(\displaystyle 3x^{2}\) is also true. \(\displaystyle 3x^{2}>lnx\) on the entire interval because at \(\displaystyle x=e\), we have

 

\(\displaystyle 3e^{2}>1=lne\). So the numerator is larger to begin with, and since:

 

\(\displaystyle 6x > 1/x\)  for all \(\displaystyle x>e\) (or any \(\displaystyle x>1\) for that matter), the derivative of the numerator is greater, too. This means the numerator will always be larger, so this condition coincides with the condition of \(\displaystyle f'(x)\) being positive.

Example Question #1602 : Functions

Find the intervals on which the following function is increasing:

\(\displaystyle f(x)=3x^2-5x+2\)

Possible Answers:

\(\displaystyle (-\infty , \frac{2}{3})\bigcup (1, \infty )\)

\(\displaystyle (\frac{5}{6}, \infty )\)

\(\displaystyle (-\infty , \infty )\)

\(\displaystyle (-\infty , \frac{5}{6})\)

Correct answer:

\(\displaystyle (\frac{5}{6}, \infty )\)

Explanation:

To find the intervals on which the function is increasing, we must find the intervals where the first derivative is positive. To do this, we must find the first derivative, and find its critical values (at which the first derivative is equal to zero):

\(\displaystyle f'(x)=6x-5\)

\(\displaystyle c=\frac{5}{6}\)

The derivative was found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Now, write the intervals of the function for which c is the upper and lower bound:

\(\displaystyle (-\infty , \frac{5}{6}), (\frac{5}{6}, \infty )\)

Note that at the critical value, the derivative is neither positive nor negative.

Now, we analyze the sign of the derivative within each interval; on the first interval, the derivative is always negative, but on the second interval, the first derivative is always positive. In other words, for this set of values - \(\displaystyle (\frac{5}{6}, \infty )\) -  the function is increasing. 

Example Question #11 : How To Find Increasing Intervals By Graphing Functions

Determint the intervals on which the following function is increasing:

\(\displaystyle f(x)=10x+6\)

Possible Answers:

\(\displaystyle (-\infty, 0)\)

\(\displaystyle (-\infty, \infty)\)

\(\displaystyle (0, \infty)\)

\(\displaystyle (-\frac{3}{5}, \infty )\)

Correct answer:

\(\displaystyle (-\infty, \infty)\)

Explanation:

To determine the intervals on which the function is increasing, we must determine the intervals on which the first derivative of the function is positive. To start, we must find the first derivative:

\(\displaystyle f'(x)=10\)

The derivative was found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

The first derivative is a positive constant, therefore the function is increasing on the entire domain, \(\displaystyle (-\infty, \infty)\).

Example Question #11 : Increasing Intervals

When is the function \(\displaystyle f(x)=x^2-4x+3\) increasing?

Possible Answers:

\(\displaystyle (2,\infty )\)

\(\displaystyle (-\infty, 2 )\)

\(\displaystyle (-\infty ,\infty )\)

\(\displaystyle (0,2)\)

\(\displaystyle (-2,0)\)

Correct answer:

\(\displaystyle (2,\infty )\)

Explanation:

To find where the function is increasing, you must first find the derivative of the function so you can test critical points. The derivative of the function is \(\displaystyle f{}'(x)=2x-4\). Then, set that equal to \(\displaystyle 0\) to find the critical points. When you set that equal to \(\displaystyle 0\), you get \(\displaystyle x=2\). Then set up a number line so you can test values to determine when the function is increasing and decreasing. We know it's changing direction at our critical point, \(\displaystyle x=2\). So let's pick a point to the left of \(\displaystyle 2\) and plug it in to the derivative. I'll pick \(\displaystyle 0\): \(\displaystyle 2(0)-4=-2\). Since the answer is negative, we know that the function is decreasing. Pick a point to the right of 2. I'll pick 3: \(\displaystyle 2(3)-4=2\). Since the answer is positive, the function is increasing. Therefore, the function is increasing from \(\displaystyle (2,\infty )\).

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